Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three unit vectors such that $|\vec{a}-\vec{b}|^2+|\vec{a}-\vec{c}|^2=8$. Then $|\vec{a}+2\vec{b}|^2+|\vec{a}+2\vec{c}|^2$ is equal to :

Numerical Value Type

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Ans. (02.00)

Sol. $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$

$|\vec{a}-\vec{b}|^2+|\vec{a}-\vec{c}|^2=8$

$\implies \vec{a}.\vec{b}+\vec{a}.\vec{c}=-2$

Now $|\vec{a}+2\vec{b}|^2+|\vec{a}+2\vec{c}|^2=2|\vec{a}|^2+4|\vec{b}|^2+4|\vec{c}|^2+4(\vec{a}.\vec{b}+\vec{a}.\vec{c})=2$

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